Problem: Simplify and expand the following expression: $ \dfrac{2}{3k + 9}+ \dfrac{3}{3k + 30}+ \dfrac{k}{k^2 + 13k + 30} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3k + 9} = \dfrac{2}{3(k + 3)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3k + 30} = \dfrac{3}{3(k + 10)}$ We can factor the quadratic in the third term: $ \dfrac{k}{k^2 + 13k + 30} = \dfrac{k}{(k + 3)(k + 10)}$ Now we have: $ \dfrac{2}{3(k + 3)}+ \dfrac{3}{3(k + 10)}+ \dfrac{k}{(k + 3)(k + 10)} $ The least common multiple of the denominators is: $ 9(k + 3)(k + 10)$ In order to get the first term over $9(k + 3)(k + 10)$ , multiply by $\dfrac{3(k + 10)}{3(k + 10)}$ $ \dfrac{2}{3(k + 3)} \times \dfrac{3(k + 10)}{3(k + 10)} = \dfrac{6(k + 10)}{9(k + 3)(k + 10)} $ In order to get the second term over $9(k + 3)(k + 10)$ , multiply by $\dfrac{3(k + 3)}{3(k + 3)}$ $ \dfrac{3}{3(k + 10)} \times \dfrac{3(k + 3)}{3(k + 3)} = \dfrac{9(k + 3)}{9(k + 3)(k + 10)} $ In order to get the third term over $9(k + 3)(k + 10)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{k}{(k + 3)(k + 10)} \times \dfrac{9}{9} = \dfrac{9k}{9(k + 3)(k + 10)} $ Now we have: $ \dfrac{6(k + 10)}{9(k + 3)(k + 10)} + \dfrac{9(k + 3)}{9(k + 3)(k + 10)} + \dfrac{9k}{9(k + 3)(k + 10)} $ $ = \dfrac{ 6(k + 10) + 9(k + 3) + 9k} {9(k + 3)(k + 10)} $ Expand: $ = \dfrac{6k + 60 + 9k + 27 + 9k}{9k^2 + 117k + 270} $ $ = \dfrac{24k + 87}{9k^2 + 117k + 270}$ Simplify: $ = \dfrac{8k + 29}{3k^2 + 39k + 90}$